Cathy's Question: Motor on/off with LDR and Resistor?

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SUMMARY

The discussion centers on designing a potential divider circuit using a Light Dependent Resistor (LDR) and a fixed resistor (R) to control a motor based on light conditions. The LDR has a resistance of 200 Ω in light and 2500 Ω in darkness. The motor activates when a voltage of 4.0 V is applied, powered by a 6.0 V battery. Calculations are required to determine the appropriate values of R (3000 Ω, 4000 Ω, and 5000 Ω) for dark conditions and (300 Ω, 400 Ω, and 500 Ω) for light conditions to achieve the necessary voltage across the motor.

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CathyLou
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I've no idea how to start the following question so any help or guidance would be really appreciated.

A potential divider circuit is set up with a LDR and a fixed resistor, R. The LDR has a resistance of 200 Ω when it is light and 2500 Ω when it is dark. A motor is set so that it comes on if 4.0 V is applied across it. A 6.0 V battery provides the emf (assume it has no internal resistance).

When it is dark, does the motor come on if R is (a) 3000 Ω (b) 4000 Ω (c) 5000 Ω?

When it is light, does the motor come on if R is (a) 300 Ω (b) 400 Ω (c) 500 Ω?

If the motor were moved so that it is across the LDR instead, what would R have to be so that the motor came on in the light?


Thank you.

Cathy
 
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need to post at least some thought about this: maybe you don't have a diagram, and that is the issue: I read it as such. The diffference between
v1 and v3 is just the applied EMF of 6Volts.

V1-----R------v2------Rx------v3. where Rx=variable depending on whether dark or light. So for the potential difference between v1 and v2, to be >=4v,
what must R be. Does this help at all?
 

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